# Does the ground state of the Schrödinger equation, in any number of dimensions, always have constant phase?

I just read this argument in this paper (PDF).

It suggests that, from variational principles, you can show that you can always lower the energy of a state by making the phase constant, thus resulting in a ground state that must have no phase change. I know this to be true for 1D systems (ground state is always real), but I'm not so convinced for the general $$N$$-dimensional case, though that's what the paper says as it uses the nabla for spatial derivatives.

My problem with their argument is that they split the trial wavefunction as

$$\psi = f(\mathbf{r})e^{i\chi(\mathbf{r})}$$

but even if you send $$\chi$$ to zero, what proof do we have that an $$f$$ that, alone, solves the Schrödinger equation always exists? Is that a separate theorem? It doesn't seem trivial to me.

• – Emilio Pisanty Sep 2 '19 at 12:15
• See also e.g. The ground state is signed and symmetric, Non-degeneracy of ground state in quantum mechanics, and other related questions on MathOverflow. – Emilio Pisanty Sep 2 '19 at 12:15
• Thanks! The first of those answers actually seems to satisfy my doubt, since it mentions that for the $n$-th state you always have a number of nodes $\leq n-1$. That leaves only 0 nodes for the ground state. – Okarin Sep 2 '19 at 12:23
• That said, I'm not sure what you're really asking here. The fact that this form of $\psi$ is the most general is obvious, and the fact that having constant $\chi$ minimizes the energy expectation value is clearly laid out in the paper. Are you asking what guarantees we have that a ground state exists and that the variational principle will find it? – Emilio Pisanty Sep 2 '19 at 12:24
• In the article it says: "Obviously, the minimization of the expectation value of the Hamiltonian within the full Hilbert space is equivalent to the exact resolution of the stationary Schrödinger equation." No assumption is made that $\psi$ solves the (time independent or stationary) Schrödinger equation. On the other hand, every function in your Hilbert state space (which consists of position dependent functions) is a valid state. The time-dependent Schrödinger equation solves for its time evolution. If I understand him correctly, this is what Javier is saying. – doetoe Sep 2 '19 at 13:59

They're not assuming that $$\psi$$ is a solution of the Schrödinger equation. What they are doing is, given an arbitrary wavefunction $$\psi$$, calculating its energy expectation value $$\langle \psi | H | \psi \rangle$$, and seeing how they can tweak $$\psi$$ to make this lower. Crucially, by "energy" they mean this mean value, and not the eigenvalue (which obviously doesn't make sense if your function is not an eigenfunction).
Now, the ground state wavefunction is the one that actually minimizes $$\langle H \rangle$$: what they're doing here is showing that given any wavefunction you can always make its mean energy $$\langle H \rangle$$ lower by removing the phase, so that the ground state wavefunction (which has the lowest mean energy of them all) must have constant phase.
I found a possible answer in this Chemistry SE answer. Apparently Courant demonstrated a theorem guaranteeing that the $$n$$-th eigenstate of the Schroedinger equation (really, of a broader class of differential equations with the same properties) always has a number of nodes $$\leq n-1$$. This inequality becomes equality only in the 1-D case; however, I think it is sufficient to answer my question, since the ground state has $$n=1$$ and thus must possess no nodes, which implies a constant phase across all space (if we take $$f$$ positive by construction).
• $\psi(x)=\exp(ix)\sin(\pi x)$ has no nodes on $(0,1)$, yet it doesn't have constant phase. So this nodal argument doesn't seem to prove anything. – Ruslan Sep 2 '19 at 12:40